A) KE
B) Zero
C) \[\frac{KE}{4}\]
D) \[\frac{KE}{2}\]
Correct Answer: C
Solution :
Let initial velocity of the particle be v. \[\therefore \] \[KE=\frac{1}{2}m{{v}^{2}}\] at the highest point of its flight \[v=u\cos {{60}^{o}}=\frac{u}{2}\] \[\therefore \] Kinetic energy at this point \[=\frac{1}{2}m{{v}^{2}}=\frac{1}{2}\frac{m{{u}^{2}}}{4}=\frac{KE}{4}\]You need to login to perform this action.
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